Prove that equation $ p^{4}+q^{4}=r^{4}$ does not have solution in set of prime numbers.
Problem
Source: Federation of Bosnia, 2. Grades 2008.
Tags: inequalities, abstract algebra, number theory, greatest common divisor, prime numbers
23.04.2008 20:12
23.04.2008 21:34
Another approach: If there is any one out of 3 variables which is divisble by 2, then it must be equal to 2. If a=2 or b=2 we easily infer that the equation doesn't have any prime solutions! If c=2 then we will have the inequality: a^4<=16 then easily to infer that a=2. But then b=0(contradicts to th assumption) Now, if there is not any variables which are divisible by 2, then a^4+b^4 module 4 is equal to 2. But c^4 module 4 is equal to 1(contradiction since a^4+b^4=c^4) From 2 cases above, we are done!
23.04.2008 23:22
Is it valid in a contest? Absolutely
23.04.2008 23:37
Quote: Fermat's last Theorem, since all primes lie in the Natural Numbers You are absolutely right but it was for 2.nd grades Maybe just a few students know about Fermat's last theorem.
24.04.2008 05:18
CatalystOfNostalgia wrote: Thus $ p^{2} = x^{2} - y^{2},q^{2} = 2xy$. The second equation gives $ q = 2,xy = 2$, so $ x = 2,y = 1$. I'm not sure how you concluded this. We can have $ x = 2a^2, y = b^2$ where $ \gcd(a, b) = 1$ and $ b$ is odd. The proof that this equation has no solution is usually done by infinite descent.
24.04.2008 05:21
t0rajir0u wrote: CatalystOfNostalgia wrote: Thus $ p^{2} = x^{2} - y^{2},q^{2} = 2xy$. The second equation gives $ q = 2,xy = 2$, so $ x = 2,y = 1$. I'm not sure how you concluded this. We can have $ x = 2a^2, y = b^2$ where $ \gcd(a, b) = 1$ and $ b$ is odd. The proof that this equation has no solution is usually done by infinite descent. You're right, t0raijru! .But it isn't essential in this case since the restriction:"prime numbers" make the problem more easier than the original problem. That maybe is suitable for the level of student in that age.
24.04.2008 13:33
I'd assume Fermat's Last Theorem wouldn't be accepted as it trivializes the problem.
24.04.2008 13:58
Quote: I'd assume Fermat's Last Theorem wouldn't be accepted as it trivializes the problem. It could be interesting if someone solved problem by Fermat's Last theorem but I assume that nobody of contestants know about it ( 2.nd grade) Most probabaly solution based on this theorem would be accepted!
25.04.2008 05:39
$ p^4=(r^2+q^2)(r+q)(r-q)$. Note that $ r^2+q^2>r+q>r-q$. Then $ r^2+q^2=p^3$, $ r+q=p$, $ r-q=1$. Then $ r=\frac{p+1}{2}$, $ q=\frac{p-1}{2}$, in the first equation, $ (p-1)(2p^2+p+1)=0$ which has no prime solution.
25.04.2008 06:05
There is a solution in Aops volume 2:
17.05.2008 02:35
A very easy approach of this problem: If p and q are odd, then r is even, but if r is even, since r is prime, r=2.Here, it's easy to see that there aren't solutions for p^4+q^4. So, at least one of this primes is even.Wlog, let p this prime, so p=2.Now, we have:16=(r²2+q²)(r²-q²) . Here, it's again easy to see that there aren't solutions.So we are done. Bye bye
17.05.2008 02:45
Maybe I'm being stupid, but just Fermats last theorem????
17.05.2008 05:17
The point is to prove this result independently. The case $ n = 4$ was known even to Fermat, and the case for $ x, y, z$ prime is even easier.
20.05.2008 06:00
1. Ok, fine. 2. This is a 2ND GRADE competition?? I can't really see how 2nd graders would be doing this.
20.05.2008 11:10
Perhaps in Bosnia, Grade 2 does not equate to Grade 2 where you're from. In California, Grade 2 is generally ages 7-8. This does seem a bit difficult for that age bracket, but again, grade 2 in Bosnia is most likely seen differently.
20.05.2008 13:26
Grade 2 (more precisely, high school 2nd grade) in Bosnia is equivalent to grade 10, i.e. 15-16 year old students.
21.05.2008 19:08
That would explain
11.03.2011 14:18
29.05.2011 19:51
1)if $p,q>2=>r=2$ no solutions)) 2)if either $p$ or $q$ equals $2=>16=(r-q)(r+q)(r^2+q^2)$ no solutions)) very easy one)
29.05.2011 21:51
t0rajir0u wrote: CatalystOfNostalgia wrote: Thus $ p^{2} = x^{2} - y^{2},q^{2} = 2xy$. The second equation gives $ q = 2,xy = 2$, so $ x = 2,y = 1$. I'm not sure how you concluded this. We can have $ x = 2a^2, y = b^2$ where $ \gcd(a, b) = 1$ and $ b$ is odd. The proof that this equation has no solution is usually done by infinite descent. Why do you think Pythagorean triples are wrong since one can take $a^2=p^4,b^2=q^4,c^2=r^4$ and get Pythagoras equation ????? That method is perfectly ok!
02.02.2018 16:14
$p, q, r\in\mathbb{P}$ $p^2=u^2–v^2$ $q^2=2xy$ not prime since $\sqrt{2}.k$ $r^2=u^2+v^2$ clearly not primes....
02.02.2018 23:22
Work in modulo 6 and the answer will be revealed.
03.02.2018 17:06
delegat wrote: Prove that equation $ p^{4}+q^{4}=r^{4}$ does not have solution in set of prime numbers. We can prove by "descente infinite" that the equation $ p^{4}+q^{4}=r^{2}$ have not positive integer solution, so the equation $ p^{4}+q^{4}=r^{4}$ have not too positive integer solutions, even more prime solutions!
26.02.2018 16:16
Yea somewhat like @rikabcd123456 says can't we simply work modulo $4$ ? $(p,q,r) \equiv \pm 1 \pmod 4 \implies p^4,q^4,r^4 \equiv 1 \pmod 4$.Then the LHS $\equiv 2 \pmod 4$ whereas RHS $\equiv 1 \pmod 4$.We are done. @targo all primes are 4k+1 or 4k+3 except 2.
26.02.2018 18:06
AYG wrote: Yea somewhat like @rikabcd123456 says can't we simply work modulo $4$ ? $(p,q,r) \equiv \pm 1 \pmod 4 \implies p^4,q^4,r^4 \equiv 1 \pmod 4$.Then the LHS $\equiv 2 \pmod 4$ whereas RHS $\equiv 1 \pmod 4$.We are done. How you got $p,q,r\equiv \pm 1\pmod{4}$ ? You need to consider that , as , $p,q<r$ , we will have one of $p$ or, $q$ equals to $2$,to maintain the parity.So, WLOG if you take $p<q$ then,$p=2$ .
26.02.2018 19:24
Is this ok? We can clearly see that $p=q=2$ is not a solution. $q^4=(r^2+p^2)(r+p)(r-p)$ So if $p\neq 2 $ we get that no part of right side is equal to $1$ and no part is negative. Thus we have that the right side is "made" of $q^2, q, q$. So one part in brackets is equal to one of this. But that would mean that 2 of them would be equal which makes no sense because we have prime numbers which are greater than $1$. So that doesn't work. Now if we have $p=2$ we just switch $p$ and $q$ and we get the same equation.
26.02.2018 19:53
delegat wrote: Prove that equation $ p^{4}+q^{4}=r^{4}$ does not have solution in set of prime numbers. Lemma: Let $\left(a,b,c\right)$ be a Pythagorean triple. Then $abc$ is a multiple of $60$. Proof: We begin by proving that one of $a$,$b$,$c$ is a multiple of $3$. Assume the contrary, then $c^2=a^2+b^2=2\left(\operatorname{mod}3\right)$. This is impossible hence one of $a$,$b$,$c$ is a multiple of $3$. Now take the numbers modulo $8$. If no $a$,$b$,$c$, then $c^2=1,4\left(\operatorname{mod}8\right)$ and $a^2+b^2=2,5,0\left(\operatorname{mod}8\right)$. Impossible, hence one of $a^2$, $b^2$, $c^2$ is a multiple of $8$ which implies one of $a$,$b$,$c$ is a multiple of $4$. Finally, take the numbers modulo $5$. Assume none are a multiple of $5$. Then $c^2=1,4\left(\operatorname{mod}5\right)$ and $a^2+b^2=2,3,0\left(\operatorname{mod}5\right)$. Hence one of $a$,$b$,$c$ is a multiple of $5$. $60=3\cdot4\cdot5$ hence $abc$ is a multiple of $60$. From the lemma, $p^2q^2r^2$ is a multiple of $60$ and hence $pqr$ is a multiple of $30$. As they are primes, the only possible solution would be $\left(p,q,r\right)=\left(2,3,5\right)$. These do not work hence there are no solutions.
28.09.2020 06:53
If $p, q, r$ are all odd primes, then there are no solutions by simply taking both sides mod 4 (LHS is $2 \pmod 4$ while RHS is $1 \pmod 4$). Hence at least one of $p, q, r$ is $2$. Clearly, it is not possible for $p=q=r=2$. It is also easy to check that we cannot have $2$ of the 3 variables be equal to 2 and the last one be an odd prime. This means that exactly one of the 3 variables must be equal to 2. If $r=2$, this is not possible as we get $p^4+q^4=16$, and the minimal value of the left side is $3^4+3^4=162 > 16$. So, WLOG let $p=2$. Therefore, $$16+q^4=r^4 \implies 16=r^4-q^4 \implies 16=(r^2+q^2)(r+q)(r-q).$$However, the minimal value of $r^2+q^2$ is $3^2+3^2=18$, so there are no solutions here. Since we've covered all the possibilities, we can conclude that there are no solutions.
10.07.2023 22:08
Fermat's last theorem and directly it has no solution
11.07.2023 01:53
Let $p, q, r$ are all $>3$ then must $p^4+q^4\equiv 2\pmod 6$ but $r^4\equiv 1\pmod 6$ done. Rest is case check of $\{2,3\}$.
21.09.2023 05:32
Groupsolved with IAmTheHazard, sixoneeight, CT17, CT18, megarnie, OronSH, MathLuis, pikapika007, Martin2001, naman12, IAteTheHazard, tienxion, Evil Chin Inc Follows immediately by Fermat's Last Theorem.
21.09.2023 10:17
P= r^2 - q^2 and P^3 = r^2 +q^2
22.09.2023 03:33
I think this was relatively simple - but maybe I did something wrong?
22.09.2023 04:03
It follows immediately by Fermat's Last Theorem. Or any 4th power of a prime is always 1 mod 3 besides if the prime is 3. Hence contradiction so one of the primes is 3. If r=3 no solutions. quick check yeilds no 4th powers have difference of 81.
22.09.2023 04:06
How do you instantly notice that any prime number to 4th power is always $1 \pmod 3$ expect for 3?
22.09.2023 05:39
PaperMath wrote: How do you instantly notice that any prime number to 4th power is always $1 \pmod 3$ expect for 3? any number to a 4th power mod 3 thats not 0 is equal to 1, since any number squared mod 3 = 1.
22.09.2023 05:41
qwerty123456asdfgzxcvb wrote: PaperMath wrote: How do you instantly notice that any prime number to 4th power is always $1 \pmod 3$ expect for 3? any number to a 4th power mod 3 thats not 0 is equal to 1, since any number squared mod 3 = 1. Uhhhhh that doesn’t answer my question. That is literally what im asking, how people can identify those relationships better. Can someone also check my solution? -
22.09.2023 06:49
The mod $3$ argument is nearly identical to what you did for mod $5$. In general, if you reduce $a^k$ mod $p$, you'll get $1$ if $p-1\mid k$. So a good place to start is with primes $p$ for which $p-1$ is a factor of the exponent.
23.09.2023 05:46
23.09.2023 16:28
hsiljak wrote: Grade 2 (more precisely, high school 2nd grade) in Bosnia is equivalent to grade 10, i.e. 15-16 year old students. I mean there are a significant amount of 10 graders that know fermat's last (like me)
23.09.2023 19:10
23.09.2023 19:59
23.09.2023 20:05
Awesome_Twin1 wrote:
Don't forget to account for the cases of $r=2,3$ as these are different!!